Ncert Solutions Maths class 12th
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The plane which bisects the line joining the points (4, −2,3) and (2,4, -1) at right angles also passes through the point:
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The plane which bisects the line joining the points (4, −2,3) and (2,4, -1) at right angles also passes through the point:
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Mid point P = (3,1,1)
Normal of plane is along the line AB.
D.R.'s of normal = 4-2, -2-4, 3-1 (-1) = 2, -6, 4
= 1, -3, 2
Plane -> 1 (x-3) - 3 (y-1) + 2 (z-1) = 0
=> x - 3y + 2z - 2 = 0
Upvote 
If a △ ABC have vertices A(−1,7), B(−7,1) and C(5, −5), then its orthocenter has coordinates:
If a △ ABC have vertices A(−1,7), B(−7,1) and C(5, −5), then its orthocenter has coordinates:
m? = 6/-12 = -1/2
∴ Equation of AD is y-7 = 2 (x+1)
y = 2x+9
m? = 12/-6 = -2
∴ Equation of BE is y-1 = 1/2 (x+7)y
y = x/2 + 9/2
by (1) and (2)
2x+9 = (x+9)/2
=> 4x+18 = x+9
=> 3x = 9 => x=-3
∴ y=3
Upvote The probability that a randomly chosen 5 - digit number is made from exactly two digits is:
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The probability that a randomly chosen 5 - digit number is made from exactly two digits is:
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Total = 9 (10? )
Fav. Way =? C? (2? -2) +? C? (2? -1) = 36 (30) + 9 (15) = 1080 + 135
Probability = (36x30+9x15)/ (9x10? ) = (4x30+15)/10? = 135/10?
Upvote A die is thrown two times and the sum of the scorers appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared at least once is
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A die is thrown two times and the sum of the scorers appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared at least once is
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Sum obtained is a multiple of 4.
A = { (1,3), (2,2), (3,1), (2,6), (3,5), (4,4), (5,3) (6,2), (6,6)}
B: Score of 4 has appeared at least once.
B = { (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (4,1), (4,2), (4,3), (4,5), (4,6)}
Required probability = P (B/A) = P (B? A)/P (A)
= (1/36) / (9/36) = 1/9
Upvote The foot of the perpendicular drawn from the point (4,2,3) to the line joining the points (1, −2,3) and (1,1,0) lies on the plane:
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The foot of the perpendicular drawn from the point (4,2,3) to the line joining the points (1, −2,3) and (1,1,0) lies on the plane:
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Equation of
AB = r = (î + j) + λ (3j - 3k)
Let coordinates of M
= (1, (1 + 3λ), -3λ).
PM = -3î + (3λ - 1)j - 3 (λ + 1)k
AB = 3j - 3k
? PM ⊥ AB ⇒ PM · AB = 0
⇒ 3 (3λ - 1) + 9 (λ + 1) = 0
⇒ λ = -1/3
∴ M = (1,0,1)
Clearly M lies on 2x + y - z = 1.
Upvote The lines
r = (î − ĵ) + l(2î + k) and
r = (2î – ĵ) + m(î + ĵ – k)
The lines
r = (î − ĵ) + l(2î + k) and
r = (2î – ĵ) + m(î + ĵ – k)
r = î (1 + 12l) + j (-1) + k (l)
r = î (2 + m) + j (m - 1) + k (-m)
For intersection
1 + 2l = 2 + m
-1 = m - 1
l = -m
from (ii) m = 0
from (iii) l = 0
These values of m and l do not satisfy equation (1).
Hence the two lines do not intersect for any values of l and m.
Upvote